Answer to Question #260850 in Linear Algebra for talie

Question

Show that the map T : R^

2 −→ R^2 defined by T(x1, x2) = (x1 −x2, 0) is a linear transformation, hence find the matrix

of this transformation relative to the basis B = {(−1, 1),(0, 1)}

Answer 1

Solution;

Given ;

$T(x_1,x_2)=\begin{bmatrix} x_1-x_2\\ 0 \end{bmatrix}$

A map T is linear if;

$T(x)=Ax$

A is the matrix of transformation;

Here;

$A=\begin{bmatrix} 1 & -1\\ 0 & 0 \end{bmatrix}$

Clearly;

$T(x_1,x_2)=\begin{bmatrix} 1 & -1\\ 0 & 0 \end{bmatrix}$ $\begin{bmatrix} x_1\\ x_2 \end{bmatrix}$ =$(x_1-x_2,0)$

Therefore the map is a linear transformation.

And;

$B=[(-1,1),(0,1)]$

Hence;

$T_B(-1,1)=\begin{bmatrix} -1-1\\ 0 \end{bmatrix}=\begin{bmatrix} -2\\ 0 \end{bmatrix}$

$T_B(0,1)=\begin{bmatrix} 0-1\\ 0 \end{bmatrix}=\begin{bmatrix} -1\\ 0 \end{bmatrix}$

Hence;

$T_B=T[(-1,1),(0,1)]=[(-2,0),(-1,0)]$

Let;

$(-2,0)=a_1(-1,1)+b_1(0,1)$

$(-2,0)=(-a_1,a_1)+(0,b_1)$

$(-2,0)=(-a_1,a_1+b_1)$

From which;

$-2=-a_1$ $\therefore a_1=2$

Also;

$O=a_1+b_1$ ;

$b_1=-a_1=-2$

Now,let;

$(-1,0)=a_2(-1,1)+b_2(0,1)$

$(-1,0)=(-a_2,a_2)+(0,b_2)$

$(-1,0)=(-a_2,a_2+b_2)$

From which;

$-1=-a_2 ,\therefore a_2=1$

Also;

$0=a_2+b_2$ ;

$b_2=-a_2=-1$

The coefficient matrix A,is ;

$A=\begin{bmatrix} 2 & -2 \\ 1 & -1 \end{bmatrix}$

Hence,the matrix of transformation of T relative to B is A^T;

$A^T=\begin{bmatrix} 2 & 1 \\ -2 & -1 \end{bmatrix}$